On $\mathbb{R}^{n+1}-\{0_{\mathbb{R}^{n+1}}\}$ define the following equivalence relation: $$x\sim y\iff\quad y=tx\;\text{for same}\; t\in\mathbb{R^{\times}}.$$
Problem. The equivalence relation $\sim$ is open.
My attempt. Let $U\subset \mathbb{R}^{n+1}-\{0_{\mathbb{R}^{n+1}}\}$ be an open set, we must prove that $\pi(U)$, where $\pi\colon \mathbb{R}^{n+1}-\{0_{\mathbb{R}^{n+1}}\}\to\big(\mathbb{R}^{n+1}-\{0_{\mathbb{R}^{n+1}}\}\big)\big/\sim$ is the projection map, is an apen set of $\big(\mathbb{R}^{n+1}-\{0_{\mathbb{R}^{n+1}}\}\big)\big /\sim$. The image $\pi(U)$ is open if and only if $\pi^{-1}(\pi(U))$ is an open set of $\mathbb{R}^{n+1}-\{0_{\mathbb{R}^{n+1}}\}$. Now \begin{split} \pi^{-1}(\pi(U))=\bigcup_{p\in U} [p]=&\bigcup_{p\in U}\big\{tp\;|\; t\in\mathbb{R}^\times\big\}\\ =&\bigcup_{t\in\mathbb{R}^\times}\big\{tp\;|\; p\in U\big\}\\ =& \bigcup_{t\in\mathbb{R}^\times} tU. \end{split} Fixed $t\in\mathbb{R}^\times$, consider the map $f_t\colon\mathbb{R}^{n+1}-\{0_{\mathbb{R}^{n+1}}\}\to\mathbb{R}^{n+1}-\{0_{\mathbb{R}^{n+1}}\}$ defined as $f_t(p):=tp.$ The map $f_t$ is an homeomorphism, then $g:=f_{|U}\colon U\to \mathbb{R}^{n+1}-\{0_{\mathbb{R}^{n+1}}\}$ is an homeomorphism too. We observe that \begin{split} g(U):=&\big\{q\in\mathbb{R}^{n+1}-\{0_{\mathbb{R}^{n+1}}\}\;\big|\;q=tp\;\text{for same}\; p\in U\big\}\\ =&\big\{tp\;\big|\; p\in U\big\}\\ =&tU. \end{split} Therefore, $U$ is open, then $g(U)$ is open, so $tU$ is open for any $t$, then $\pi^{-1}(\pi(U))$ is open.
Question. It's correct?
Thanks!
You could have shortened the proof that $tU$ is open : $tU$ is the set of thoses $x$ such that $x/t$ is in $U$, but $x\mapsto x/t$ is a continuous function. Otherwise it is correct.